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I have a doubt, i hope not so stupid.

Suppose we consider a buffer solution of acetic acid/acetate at pH = pKa = 4.76 and we add aspirin (pKa = 3.5): given that the pH of the solution is higher than the pKa of the drug, then the aspirin prevails in its ionized form, from Henderson-Hasselbalch equation. That said, does the degree of dissociation of aspirin actually depend on the pH of the solution, or does it depend on the ability of the aspirin to react with the conjugate base of the buffer to some extent?

AspH + CH3COO- <=> Asp- + CH3COOH

In fact, given that the pKa of aspirin is higher than that of acetic acid, the reaction is shifted to the right, so that aspirin will be dissociated in a certain percentage which depends precisely on this equilibrium, rather than on the pH of the solution (?).

Similar doubt regarding an unbuffered solution: suppose we want to extract aspirin from an organic solvent, with a saturated bicarbonate solution: the reaction between the two species is, also in this case, shifted to the right, as aspirin is a stronger acid than carbonic acid, therefore aspirin will prevail in the deprotonated form and there is a certain degree of dissociation. Again, does the degree of dissociation of aspirin actually depend on this reaction, or on the pH that is reached AFTER the reaction, again using the Henderson-Hasselbalch equation?

Thanks.

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  • $\begingroup$ How much acetic acid, sodium acetate and aspirin do you dissolve in your water ? $\endgroup$
    – Maurice
    Commented Mar 8, 2023 at 11:04
  • $\begingroup$ pH=pKa, so [CH3COOH]=[CH3COO-]=0,5 M for example. About aspirin, every quantity. I think that this Isn't much important (?), I Just wanted to know the base principle about what i wrote. $\endgroup$
    – Luckenberg
    Commented Mar 8, 2023 at 11:58
  • $\begingroup$ The proton exchange between aspiring and acetic acid respective forms is just an equilibrium dependent of the 2 dissociation equilibrii. The pH given by the acetic buffer ( in excess wrt aspirin) determines [A-]/[HA] ratios for both. It is like K1 = x.y1/z1 and K2 = x . y2/z2, where x = [H+]. $\endgroup$
    – Poutnik
    Commented Mar 8, 2023 at 13:00
  • $\begingroup$ @Poutnik so Is It the pH of the solution that determines the degree of dissociation, and not directly the reaction between aspirin and acetate? Consequently, is the same also true in the case of bicarbonate? That is, is it the final pH, after the reaction between aspirin and bicarbonate, that determines the degree of dissociation of the aspirin itself, and not the reaction itself? $\endgroup$
    – Luckenberg
    Commented Mar 8, 2023 at 13:40
  • $\begingroup$ It depends what has the "upper hand", if the solute presence affects/determines pH or if pH affects/determines solute behavior. In our case, the PH buffer and therefore pH has the upper hand. $\endgroup$
    – Poutnik
    Commented Mar 8, 2023 at 14:43

1 Answer 1

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(Partial answer yet)

If we take the H-H equation:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {\frac{[\ce{A-}]}{[\ce{HA}]}}$$

we can construct this:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a,Asp} + \log {\frac{[\ce{Asp-}]}{[\ce{AspH}]}} = \mathrm{p}K_\mathrm{a,\ce{CH3COOH}} + \log {\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$$

$$\mathrm{p}K_\mathrm{a,Asp} - \mathrm{p}K_\mathrm{a,\ce{CH3COOH}} = \log {\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}} - \log {\frac{[\ce{Asp-}]}{[\ce{AspH}]}}$$

$$\mathrm{p}K_\mathrm{a,Asp} - \mathrm{p}K_\mathrm{a,\ce{CH3COOH}} = \log {\left(\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} \cdot \frac{[\ce{AspH}]}{[\ce{Asp-}]}\right)}$$

$$10^{\left(\mathrm{p}K_\mathrm{a,Asp} - \mathrm{p}K_\mathrm{a,\ce{CH3COOH}}\right)} = \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} \cdot \frac{[\ce{AspH}]}{[\ce{Asp-}]}$$

As $\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}$ is given by the acetate $\mathrm{pH}$ buffer, $\frac{[\ce{AspH}]}{[\ce{Asp-}]}$ is given too.

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  • $\begingroup$ Thanks a lot for the answer. So from this discussion I understand that it is the pH that influences the degree of dissociation and not the reaction between aspirin and acetate which would take place, in the case of a buffer. But in the case of the saturated bicarbonate solution (unbuffered) what happens? Is it the reaction that takes place between bicarbonate and aspirin that directly influences the degree of dissociation or is it the final pH, which is obtained as a result of the reaction, that influences it? $\endgroup$
    – Luckenberg
    Commented Mar 9, 2023 at 15:52
  • $\begingroup$ Well, bicarbonate is a kind of pH buffer too. $\endgroup$
    – Poutnik
    Commented Mar 9, 2023 at 16:11
  • $\begingroup$ In theory, in order to form a buffer, it is necessary that the concentrations of acid and conjugate base are close enough. When you create a saturated solution of bicarbonate, tiny amounts of carbonic acid and carbonate are formed, so it's not really a buffet. To make it easier, I'm assuming that I use NaOH instead of bicarbonate to extract the aspirin from an organic solvent. Returning to the previous question: on what does the degree of dissociation depend in this case? pH can change now. $\endgroup$
    – Luckenberg
    Commented Mar 9, 2023 at 19:06
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    $\begingroup$ pH of acidic salts of biprotic acids is approximately the average of both pKa, with the capacity depending on pKa difference. $\endgroup$
    – Poutnik
    Commented Mar 9, 2023 at 19:51

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